Let $K$ be a finite field with $q$ elements.
Show that if $U$ is the subgroup of $Aut(K(x)/K)$ which consists of all mappings $\sigma$ of the form $(\sigma \theta)(x) = \theta(ax+b)$ with $a \neq 0$ then $F(U) = K((x^q -x)^{q-1})$.
I am not getting any clue to solve the problem. Help Needed.
Here $F(U)$ is the fixed field of $U$. $K(X)$ is the field of fractions.
The elements of $K((x^q -x)^{q-1})$ are of the form $$\frac{f((x^q -x)^{q-1})}{g((x^q -x)^{q-1})}.$$
$F(U) = \{ \frac fg \in K(x) \mid \sigma(\frac fg) = \frac fg \ \ \ \forall \sigma \in U\}$.
Hints are also welcome. Thank You.
Lemma: Let $K$ be a field, and $G\subseteq Aut (K(x)/K)$ a finite subgroup of order $n$. Then the polynomial $$p(T):=T^n+a_{n-1}T^{n-1}+\ldots+a_1T+a_0 =\prod_{\sigma \in G}(T-\sigma x)$$ is defined over the fixed field $K(x)^G$, and for any $i\in \{0,1,\ldots,n-1\}$ such that $a_i\notin K$, we have $K(x)^G=K(a_i).$
Proof: Note that the $\pm a_i$'s are given by the elementary symmetric polynomials (in $n$ variables) evaluated at $(\sigma_1x,\ldots,\sigma_n x)$. Thus $\sigma a_i=a_i$ for all $\sigma \in G$, which gives $p(T)\in K(x)^G[T ]$. Also, since $x$ is transcendental over $K$ and $p(x)=0$, there must exist $i\in \{0,1,\ldots,n-1\}$ such that $a_i\notin K$. For any such $a_i$, we have $a_i=g(x)/f(x)$, where $f, g\in K[x]$ are polynomials of degree $\leq n$. Therefore, $g(T)-a_if(T) \in K(a_i)[T]$ is a polynomial of degree $\leq n$ vanishing at $x$, and then $[K(x):K(a_i)]\leq n$. Now, since $K(a_i)\subseteq K(x)^G \subseteq K(x)$, and $[K(x):K(x)^G]=|G|=n$ (Artin's thm), it follows that $K(a_i)=K(x)^G$. q.e.d.
Now, for $K:=\mathbb{F}_q$ and $G:=\{x\mapsto ax+b|(a,b)\in K^*\times K\}\subseteq Aut(K(x)/K)$, we have
$$p(T)=\prod\limits_{\sigma \in G}(T-\sigma(x))=\prod\limits_{(a,b)\in K^*\times K}(T-(ax+b)).$$ Since $$a_0:=p(0)=\prod\limits_{(a,b)\in K^*\times K}(-(ax+b))=-(x^q-x)^{q-1} \notin K,$$ the previous Lemma gives $K(x)^G=K((x^q-x)^{q-1})$.